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Suppose we have an array of size n, we have to find the minimum number of moves required to make all array elements the same, where a move means incrementing n - 1 elements by 1.So, if the input is like [3, 2, 3, 4], then the output will be 4.To solve this, we will follow these steps −n := size of numsif n is same as 0, then −return 0sort the array numsans := 0for initialize i := 0, when i < n, update (increase i by 1), do −ans := ans + nums[i] - nums[0]return ansExample Let us see ... Read More
Suppose we have n points in the plane that are all pairwise distinct. Now a "boomerang" is a tuple of points like (i, j, k) such that the distance between i and j is the same as the distance between i and k. We have to find the number of boomerangs.So, if the input is like [[0, 0], [1, 0], [2, 0]], then the output will be 2, as two boomerangs are [[1, 0], [0, 0], [2, 0]] and [[1, 0], [2, 0], [0, 0]].To solve this, we will follow these steps −counter_of_boomerangs := 0for each point_1 in points array, ... Read More
Suppose we have a string s. We have to count the number of segments in a string, where a segment is defined to be a contiguous sequence of characters (no whitespace).So, if the input is like "Hello, I love programming", then the output will be 4, as there are 4 segments.To solve this, we will follow these steps −n := 0for initialize i := 0, when i < size of s, update (increase i by 1), do −if s[i] is not equal to white space, then −(increase n by 1)while (i < size of s and s[i] is not equal ... Read More
The various types of locators that Selenium support are listed below −ID − This attribute is unique for every element.Syntax − driver.findElement(By.id("")).Name − This attribute is not unique for every element.Syntax − driver.findElement(By.name("")).CSS Selector − This can be derived from element tags and attributes.Syntax −driver.findElement(By.cssSelector("")).xpath − This can be derived from element tags and attributes.Syntax −driver.findElement(By.xpath("")).TagName − This can be derived from the HTML tags to identify the elements.Syntax − driver.findElement(By.tagName("")). LinkText − This can be derived from the anchor text to identify the elementszSyntax − driver.findElement(By.linkText("")).PartialLinkText − This can be derived from the partial anchor text to identify the elementsSyntax − driver.findElement(By.partialLinkText("")).Classname ... Read More
Suppose we have a non-empty array of integers; we have to find the third maximum number in this array. If there is no 3rd max number, then return the maximum one. The challenge is, we have to solve this using linear time complexity.So, if the input is like [5, 3, 8, 9, 1, 4, 6, 2], then the output will be 6.To solve this, we will follow these steps −initialize a, b, c with NULLfor initialize i := 0, when i < size of nums, update (increase i by 1), do −if a is null or nums[i] >= value of ... Read More
Suppose we have a string which consists of lowercase or uppercase letters, we have to find the length of the longest palindromes that can be built with those letters. Now the string is case sensitive, so "Aa" is not considered a palindrome here.So, if the input is like "abccccdd", then the output will be 7, as one longest palindrome that can be built is "dccaccd", whose length is 7.To solve this, we will follow these steps −Define one map mpfor each character i in s(increase mp[i] by 1)ma := 0, c := 0, ans := 0for each key-value pair i ... Read More
The total number of headers in the web table can be counted with the help of findElements() method. The logic is to return a list of web elements with xpath with the help of tag inside the table, then getting the size of that list.Code ImplementationExampleimport org.openqa.selenium.By; import org.openqa.selenium.Keys; import org.openqa.selenium.WebDriver; import org.openqa.selenium.WebElement; import org.openqa.selenium.chrome.ChromeDriver; import java.util.concurrent.TimeUnit; import java.util.List; public class TableHeaderCount { public static void main(String[] args) { System.setProperty("webdriver.chrome.driver", "C:\Users\ghs6kor\Desktop\Java\chromedriver.exe"); WebDriver driver = new ChromeDriver(); String url = "https://www.tutorialspoint.com/plsql/plsql_basic_syntax.htm"; driver.get(url); ... Read More
Suppose we have an integer; we have to devise an algorithm to convert it to hexadecimal. For negative numbers we will use the two’s complement method.So, if the input is like 254 and -12, then the output will be fe and fffffff4 respectively.To solve this, we will follow these steps −if num1 is same as 0, then −return "0"num := num1s := blank stringwhile num is non-zero, do −temp := num mod 16if temp
The total number of links in a page can be counted with the help of findElements() method. The logic is to return a list of web elements with tagname anchor, then getting the size of that list.Code ImplementationExampleimport org.openqa.selenium.By; import org.openqa.selenium.Keys; import org.openqa.selenium.WebDriver; import org.openqa.selenium.WebElement; import org.openqa.selenium.chrome.ChromeDriver; import java.util.concurrent.TimeUnit; public class LinkCount { public static void main(String[] args) { System.setProperty("webdriver.chrome.driver", "C:\Users\ghs6kor\Desktop\Java\chromedriver.exe"); WebDriver driver = new ChromeDriver(); String url = "https://www.tutorialspoint.com/index.htm"; driver.get(url); driver.manage().timeouts().implicitlyWait(12, TimeUnit.SECONDS); //Using tagname with anchor List ... Read More
Suppose we have two strings s and t which consist of only lowercase letters. Now, string t is generated by random shuffling string s and then add one more letter at a random index. We have to find the letter that was added in t.So, if the input is like "mnopq", "pqmnot", then the output will be "t", this is the extra letter.To solve this, we will follow these steps −sSum := 0, tSum := 0for initialize i := 0, when i < call length() of s, update (increase i by 1), do −sSum := sSum + s[i]for initialize j ... Read More